The correct option is A (2n+1)π2,n∈Z
Given that |z1+z2|=|z1−z2|
⇒|z1+z2|2=|z1−z2|2
[∵|z1−z2|2=|z1|2+|z2|2−2Re(z1¯¯¯¯¯z2 ),|z1+z2|2=|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )]
⇒|z1|2+|z2|2+2Re(z1¯¯¯¯¯z2 )=|z1|2+|z2|2−2Re(z1¯¯¯¯¯z2 )
⇒4Re(z1¯¯¯¯¯z2 )=0
⇒Re(z1¯¯¯¯¯z2 )=0
Let z1=r1eiθ1 and z2=r2eiθ2
Then z1¯¯¯¯¯z2=r1r2eiθ1e−iθ2=r1r2ei(θ1−θ2)
⇒z1¯¯¯¯¯z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))
∵Re(z1¯¯¯¯¯z2 )=0
∴cos(θ1−θ2)=0
⇒θ1−θ2=(2n+1)π2,n∈Z